## Friday, 26 August 2016

### Evoultion of Wave Function - Splitting the Box

I was struggling with the structure of SchrÃ¶dinger's equation over time when we consider a real physical process. The problem can be stated in two parts as follows,

Is it possible to construct a wave function for a system with a new introduced potential from the previously known wave function solution of the same system before the introduction of the potential.  Eg: I want to get the solution of an infinite square well where the dirac delta potential is placed at the center from the usual square well problem without potential. We are not talking about entirely two things. Basically, it is a process of introducing an infinite square well and slowly putting the delta potential at the center over time.

The second part is where we don't introduce any kind of bizarre potentials, instead we change the boundary conditions over time. It is like expanding the square well or making some regions in space as inaccessible to the wave function.

Note:

There is a difference between wave function going to zero at some place and the space itself becomes inaccessible to the wave function.

One should not to be confused these things with the concept of perturbation even though it looks similar in some places. In perturbation, there is no change in boundary conditions or any introduction of completely arbitrary potentials. It is a very small change in the initial potential with same conditions maintained at

all the time.

To understand this, I took a series of problems with different levels in ascending order. First, we try to solve the general schrodinger equation for one dimensional infinite square well with a very little change by moving it to length "a" and "2a". Except for the limits, the solution is written in normal convention of "k" and "E" , we have $$\psi = A e^{ikx} + B e^{-ikx}$$  and applying the left, right limits, $$A e^{ika} + Be^{-ka} = 0 \,\,\,...(1)\\ A e^{i2ka} + B e^{-i2ka}\,\,\,...(2)$$ (1) gives $$A e^{i2ka}= -B$$ provided $e^{ika}$ don't become zero. $$(2) \rightarrow \,\,\, B (e^{-i2ka}-1) = 0 \\ B = 0 \,\,or\,\, e^{-2ika} = 1$$  B = 0 gives A=0 i.e. trivial solution where there is no wave function. It leaves us with $$e^{-i2ka} = 1 \\ A = -B\\ cos2ka = 1 \,\,and\,\, sin 2ka = 0$$both conditions need to be satisfied if, $$sin 2ka = 0 \,\,and \,\,cos2ka = 1 \\ 2ka = n\pi \,\,and\,\, 2ka = 2n\pi$$ larger condition is the common one i.e. $$ka = n\pi$$ which gives us the exact solution $$\psi = A e^{ika} - A e^{-ika} = C sin 2ka$$ where "C" is again an overall constant to be determined by normalization and $ka = n\pi$ Normalization gives again the same constant [ since, $$sin^2 2kx = \frac{1 - cos2kx}{2}$$ where integration is from "a" to "2a" i.e. sin 2ka = sin ka = 0 ] i.e. $\sqrt{\frac{2}{a}}$ giving us the complete solution,$$\psi (x) = \sqrt{\frac{2}{a}} sinkx \,\,\,\,and\,\, k = \frac{n\pi}{a}$$

It was quite a bit surprise for me, that there is no any kind of reflection about the shift of the coordinate system or the shift about the coordinate syystem. Even though the wave function depends on the position, it doesn't reflect anything i.e. independent of coordinate system.

But don't think they are exactly the same in every way because now the limits have changed to give negative values for all x values in between "a" and "2a". But, it doesn't matter since the measurable quantities depend on the square of the wave function - saves the day.

With this information, if I take the a box of length "2a" and split it into two parts "0" to "a" and "a" to "2a" with an imaginary line drawn at x = a . Actually, there is no anything physical about this problem, but I just wanted to see how the problem will evolve from one to two.

As we have two parts, the solution is, $$\psi(x) = Ae^{ikx} + Be^{-ikx} \,\,\,\,0\leq{x}\leq{a} \\ Ce^{ikx} + De^{-ikx} \,\,\,\,{a}\leq{x}\leq{2a}$$ with the boundary condition that the first part should vanish at the left x= 0 and the right should vanish at x=2a, we have, $$A+B = 0 \\ C e^{2ika} + De^{-i2ka} = 0 \\ \rightarrow \,\,\,\, Ce^{i4ka} = -D$$

If we seek the continuity at x = a, by comparing independent sine and cosine terms $$A - B = C - D \\ (C+D) coska = 0$$ Putting all the conditions together and working out gives, $$A = C \\ B = D \\ A= -B\\ k = \frac{n\pi}{2a}$$ which gives the wave function as, $$\psi(x) = F sin(kx)$$ After Normalizing, $$\psi(x) = \frac{1}{\sqrt{a}} sin(kx) \,\,\,\,or\\ \frac{1}{\sqrt{a}} i sin(kx) \\ k = \frac{n\pi}{2a} \,\,\,\, n= 0, \pm{1},\pm{2},..$$

I put 'i' because it is the exact solution I got and I just wanted to show "i" doesn't affect our wave function in anyway. The physical results are the same.

Actually we can talk a little about this 'i' . Any wave function, even though it is normalized, there is no unique form for its' structure. It is always possible to multiply it by 'i' or '-i' like we can do it in classical sense, multiplying by '1'. Hence, there is no way of telling whether your wave function solution is real or complex by just looking at it.

This itself is a justification for why the modulus square of the wave function that all matters for any physical result (which is called the probability density) and not the value of the wave function (probability amplitude).

## Monday, 23 May 2016

### Yang Mills Theory - Part - 3

We have our Lagrangian $$L = ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi - mc^2\tilde\psi\psi$$
After the gauge transformation defined as, $$\psi' = S\psi \\ where\,\,S = e^{-i\frac{q}{c\hbar}\vec{\sigma}\cdot\vec{\lambda(x)}}$$ $$L' = ic\hbar\tilde\psi'\gamma^\mu\partial(S\psi) \\= ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}\partial_\mu\psi + ic\hbar\tilde\psi{S^\dagger}\gamma^\mu(\partial_\mu{S})\psi$$gives $$L' = \left[ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi = L \right]+ {extra\,term}$$[assumed $\gamma^\mu$ commutes with S]

To compensate this extra term we use the previous mathematical tool, i.e. we define the covariant derivative as, $$D_\mu = \partial_\mu + \frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A}_\mu$$
where the new $\vec{A_\mu}$ term should cancel our previous extra terms on transformation. [The new term "$\vec{A_\mu}$" now has three components as to cancel each component in $\vec{\lambda(x)}$]

The purpose of this new covariant derivative is,

If we define our Lagrangian using this derivative as $$L = ic\hbar\tilde\psi\gamma^\mu{D_\mu}\psi$$ then we should have the same Lagrangian after the transformation $$L' = ic\hbar\tilde\psi'\gamma^\mu{D'}_\mu(S\psi) = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}(D_\mu\psi) = ic\hbar\tilde\psi\gamma^\mu{D_\mu\psi}$$ where we implied the condition, $$D'_\mu(S\psi) = S(D_\mu\psi)$$Remember, unlike the usual derivatives, Covariant derivatives also transform under the gauge transformation.

From our condition, the transformation rule for $\vec{A_\mu}$ is derived as, $$D_\mu' \psi'=\left(\partial_\mu+\frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A'_\mu}\right)(S\psi) = S \left[\left(\partial_\mu+\frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A_\mu}\right)\psi\right]$$ Using the previous relations, $$\left(\partial_\mu{S}\right)\psi + S(\partial_\mu\psi) + \frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A'_\mu}(S\psi) = S(\partial_\mu\psi) + \frac{iq}{c\hbar}S\left(\vec{\sigma}\cdot\vec{A_\mu}\psi\right)$$ Cancelling the terms and taking out psi $$\left[(\partial_\mu{S}) +\frac{iq}{c\hbar}\left(\vec{\sigma}\cdot\vec{A'_\mu}\right)S\right]\psi=\left[\frac{iq}{c\hbar}S\left(\vec{\sigma}\cdot\vec{A_\mu}\right)\right]\psi$$ Multiplying with the $S^{-1}$ operator on the right hand side we get the transformation relation, $$\vec{\sigma}\cdot\vec{A'_\mu} = S\left(\vec{\sigma}\cdot\vec{A_\mu}\right)S^{-1}+ \frac{ic\hbar}{q}(\partial_\mu{S})S^{-1}$$
Now, we can check the invariance of the Lagrangian, $$L_{new} = ic\hbar\tilde\psi\gamma^\mu(D_\mu\psi) = ic\hbar\tilde\psi\gamma^\mu\left(\partial_\mu + \frac{iq}{\hbar{c}}(\vec{\sigma}\cdot\vec{A_\mu})\right)\psi$$ After the transformation, $$L'_{new} = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu(D'_\mu(S\psi))\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[\partial_\mu(S\psi)+\frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A'_\mu})(S\psi)\right] \\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[S\partial_\mu\psi+(\partial_\mu{S})\psi + \frac{iq}{c\hbar}S(\vec{\sigma}\cdot\vec{A_\mu})S^{-1}S\psi+\frac{iq}{c\hbar}\frac{ic\hbar}{q}(\partial_\mu{S})S^{-1}(S\psi)\right]\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[S(\partial_\mu\psi)+(\partial_\mu{S})\psi+\frac{iq}{c\hbar}S(\vec{\sigma}\cdot\vec{A_\mu})\psi - (\partial_\mu{S})\psi\right]\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}\left[\partial_\mu\psi + \frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A_\mu})\psi\right]$$Gives finally the same, $$L'_{new} = ic\hbar\tilde\psi\gamma^\mu\left(\partial_\mu+\frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A_\mu})\right)\psi = L_{new}$$ Thus, we can check the invariance of our new Lagrangian.

Now, we expand our definition of S by the assumption $\lambda$ is very small so that all higher order terms can be neglected. After this approximation, substitution of the new "S" in our transformation equation for $A'_\mu$ yields, $$\vec{\sigma}\cdot\vec{A'_\mu} \approx \vec{\sigma}\cdot\vec{A_\mu} + \vec{\sigma}\cdot\partial_\mu\lambda + \frac{iq}{c\hbar}\left[(\vec{\sigma}\cdot\vec{A_\mu}),(\vec{\sigma}\cdot\vec{\lambda})\right]$$ Using the rule, $$(\vec{\sigma}\cdot\vec{a})(\vec{\sigma}\cdot\vec{b}) = \vec{a}\cdot\vec{b} + i\sigma\cdot(\vec{a}\times\vec{b})$$ we get, $$\vec{\sigma}\cdot \vec{A'_\mu} =\vec{\sigma}\cdot\vec{ A_\mu} +\vec{\sigma}\cdot\partial_\mu\vec{\lambda} + \frac{iq}{c\hbar}\left(2i\vec{\sigma}\cdot (\vec{A_\mu}\times\vec{\lambda})\right)$$gives$$\vec{A_\mu'} = \vec{A_\mu} + \partial_\mu{\vec{\lambda}}+\frac{2q}{c\hbar}(\vec{\lambda}\times\vec{A_\mu})$$ Thus, we got our new Lagrangian and the new transformation rules for corresponding terms.

But, from the same argument for the vector potential in previous problem invokes its field term in our Lagrangian. Once again, we take the Proca Lagrangian term without mass to explain our vector potential term.

You can ask, why we can't take the Proca Lagrangian with mass and redefine our $A'_\mu$ such that the invariance holds for $$A^\nu{A_\nu}$$ Yes. But we have already just finished defining the transformation rules for $A'_\mu$ and it doesn't have the invariance. Maybe, if you can find $A'_\mu$ such that, it satisfies both the above transformation and the invariance of $A^\nu{A_\nu}$ it would be wonderful. To make a comment, I should try and workout completely and find the reason for the impossibility (if there is any).

Not even $A^\nu{A_\nu}$ but also the $F^{\mu\nu}F_{\mu\nu}$ term is not invariant when we redefine our $F_{\mu\nu}$ for three vector potentials as, $$L_{extra} = \frac{-1}{16\pi}F^{\mu\nu}_1F^1_{\mu\nu}-\frac{1}{16\pi}F^{\mu\nu}_2F^2_{\mu\nu}-\frac{-1}{16\pi}F^{\mu\nu}_3F^3_{\mu\nu} = \frac{-1}{16\pi}\vec{F}^{\mu\nu}\vec{F}_{\mu\nu}$$ We will separately deal with its transformation rules and how it can be restated with additional terms in the next post.

Reference: Introduction to Elementary Particle Physics - Griffiths

### Yang Mills Theory - Part - 2

We previously constructed our Lagrangian in a simplified form using the matrix notation for the combined spinor field.

$$L = ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi - mc^2 \tilde\psi\psi$$ But, now we cannot define transformation as previous, since the new $\psi$ here is a column matrix. The transformation should be redefined in the matrix representation. So, we introduce an unitary operator U acting on our wave function to represent gauge transformation. $$\psi' = U\psi \\ \tilde\psi' = \tilde\psi U^\dagger \\ where U^\dagger{U} =\mathbb{I}$$

For this transformation, the Lagrangian is invariant, if the $\gamma^\mu$ matrices commute with the Unitary matrix.

To apply the same case for Local gauge invariance, where we used to get an extra term, we need a new representation and Unitary matrix in general can be represented using Hermitian matrices as, $$U = e^{iH}$$ where a general Hermitian matrix is defined using four parameters using Pauli matrices, $$H = a_0\mathbb{I} + a_1\sigma_1 + a_2\sigma_2 +a_3\sigma_3 = a_0\mathbb{I}+a\cdot\sigma$$ where $a\cdot\sigma = a_1\sigma_1+a_2\sigma_2+a_3\sigma_3$ in shorthand notation.

Thus, we get our Unitary matrix, $$U = e^{ia_0\mathbb{I}} e^{ia\cdot\sigma}$$ where $e^{ia\cdot\sigma}$ has determinant "1" and corresponds to SU(2).

With the same strategy as used in previous, we define $$\lambda(x) = - \frac{\hbar{c}}{q} \vec{a}$$
actually it is not a vector in the usual sense. But, the notation is preferred here to differentiate it from others.

So, we have, $$\psi' = e^{-\frac{q}{c\hbar}\vec{\sigma}\cdot\vec{\lambda(x)}}\psi$$ "$e^{ia_0\mathbb{I}}$" being a simple phase factor doesn't affect the final result even though it is a function of the coordinates.

For eg: We redefine $$U = e^{ia_0\mathbb{I}}e^{ia\cdot\sigma} = e^{ia_0}S$$ where "S" is also unitary matrix and Identity is implied in the exponential of $a_0$. Under the unitary transformation.
$$L' = i\hbar{c}\tilde\psi'\gamma^\mu\partial_\mu\psi'$$
where $mc^2\tilde\psi'\psi$ term doesn't affect the invariance in both local and global gauge tranformation. So, we have, $$L' = ic\hbar\tilde\psi{U^\dagger}\gamma^\mu\partial_\mu(U\psi)$$ $$L' = ic\hbar\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu\partial_\mu(e^{ia_0}S\psi)$$ Expanding the differential (call $ic\hbar = k)$ , $$L' = k \tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu\left( (\partial_\mu{e}^ia_0)S\psi + e^{ia_0}\partial_\mu(S\psi)\right)$$ $$L'= k\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu{i}e^{ia_0}\partial_\mu(a_0)(S\psi) + k\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu{e^{ia_0}}\partial_\mu(S\psi)$$ Defining $S\psi$ as our new $\psi'$ we have $\psi' = S\psi$ we have, $$L = ic\hbar\tilde\psi'\gamma^\mu{i}(\partial_\mu(a_0))\psi' + ic\hbar\tilde\psi'\gamma^\mu\partial_\mu(\psi')$$
with the same definition of $\lambda$, $$L' = -q\tilde\psi'\gamma^\mu\psi'\partial_\mu\lambda(x) + ic\hbar\tilde\psi'\gamma^\mu\partial_\mu\psi'$$ where the first term is exactly what we got as additional term in the previous lagrangian for which we have already defined transformation rules
[It has its own lagrangian and own vector potential - $A_0$].
So, we just need to focus on the second term.

Thus, we have our new transformed Lagrangian as ,$$L' = ic\hbar\tilde\psi'\gamma^\mu\partial_\mu\psi' - mc^2\tilde\psi\psi$$
We will focus entirely on this new Lagrangian in the next post.

Reference: Introduction to Elementary Particle Physics - Griffiths